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Proof step 90 from proof shiftright-rec-8-neg (previous step)

(i - 1) / 2 = j0 * 256, j0 * 256 >> (j - 1) = j0 >> (j - 9), 8 < j, i < 0, 0 < j, j0 < 0, ∀ i, j0. i→n(j0) < i→n(j) → 8 ≤ j0 ∧ i < 0 ∧ 0 < j0 ∧ 0 ≤ j0 → i >> j0 = (i - 255) / 256 >> (j0 - 8) ⊦ j = 0, j0 >> (j - 9) = (i / 512 - 1) >> (j - 9), j = 8, i % 256 = 0, i / 512 = 1 + j0

User applied rule insert elim lemma

And obtained the following new goals

Proof step 91:
i - 1 = i0 * 2 + j1 ∧ j1 ≤ 0 ∧ abs(j1) < 2 ∧ i0 ≤ 0 ∧ 0 < 2 ∧ i - 1 < 0, i0 = j0 * 256, j0 * 256 >> (j - 1) = j0 >> (j - 9), 8 < j, i < 0, 0 < j, j0 < 0, ∀ i, j0. i→n(j0) < i→n(j) → 8 ≤ j0 ∧ i < 0 ∧ 0 < j0 ∧ 0 ≤ j0 → i >> j0 = (i - 255) / 256 >> (j0 - 8) ⊦ j = 0, j0 >> (j - 9) = (i / 512 - 1) >> (j - 9), j = 8, i % 256 = 0, i / 512 = 1 + j0

Rule argument

seq: ⊦ 0 < j ∧ i < 0 → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i0 ≤ 0) subst: [i, j, i0, j0] → [i - 1, 2, i0, j1] rotate: T spec: int-div inst: name: div-elim-negpos

Additional information

Used simplifier rules: ⊦ i - j < i0 ↔ i < j + i0;% ⊦ i < 1 ↔ i ≤ 0;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i < j → ¬ j < i;%⊦ 0 < j ∧ i < 0 → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i0 ≤ 0); (elim) Used terms: i - 1, 2, i0, j1